Half controlled single bridge rectifier

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    Bridge Rectifier
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Discussion Overview

The discussion revolves around the operation of a half controlled single bridge rectifier supplying an R-L load, specifically focusing on the fraction of the cycle during which the freewheeling diode conducts. Participants explore various interpretations of the circuit and the role of the diodes involved.

Discussion Character

  • Homework-related
  • Debate/contested
  • Technical explanation

Main Points Raised

  • One participant suggests that each diode conducts for half the cycle, leading to an initial conclusion that the fraction of conduction is 1/2, while the book states it is alpha / pi.
  • Another participant questions the absence of a freewheeling diode in the original post, indicating confusion about the circuit diagram presented.
  • Clarifications are made regarding which diodes are considered freewheeling diodes, with some participants identifying D1 and D2 as such.
  • A participant argues that if the load current is continuous, the diodes must share conduction equally, supporting the original claim that the fraction is 1/2.
  • Further elaboration is provided on the operation of the diodes during different intervals of the cycle, leading to the conclusion that the freewheeling diode conducts for a duration of alpha in each half cycle, resulting in the book's answer of alpha / pi.
  • One participant expresses a preference for their initial explanation over the book's reasoning, indicating a subjective view on the clarity of the explanations provided.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the interpretation of the freewheeling diode's role and the correct fraction of the cycle it conducts. Multiple competing views remain regarding the definitions and explanations of the diodes involved.

Contextual Notes

There is ambiguity regarding the definitions of freewheeling diodes and the specific operation of the circuit under different conditions. The discussion reflects differing interpretations of the circuit diagrams and the assumptions made about diode conduction.

jaus tail
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Homework Statement


A half controlled single bridge rectifier is supplying an R-L load. It is operated at firing angle 'alpha'. Load current is continous. What is fraction of cycle that the freewheeling diode conducts?

Homework Equations


upload_2016-12-1_14-52-25.png

The Attempt at a Solution


Well I guess each diode will conduct for half the cycle like from 0 to pie and then from pie to 2 pie, the other diode will conduct.
So the answer must be 1/2. But book answer is alpha / pie.[/B]
 

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None of the 4 devices shown in your OP is a free-wheeling diode. Why is there no such diode shown in a question that apparently refers to it? :oldconfused:
 
I thought FWD are the diodes D1 and D2. As per first diagram in the link:
http://www.technik-emden.de/~elmalab/projekte/ws9899/pe_html/ch06s1/ch06s1p1.htm

and book diagram:
upload_2016-12-1_17-7-20.png
 
FWD in your attachment is a freewheeling diode.
 
So it's a diode parallel to RL load with anode of diode towards up and cathode facing down.
Thanks.
 
If load current is continuous, then the diodes in the rectifier must share conduction equally and your original answer would have to be right.

There is "freewheeling" operation due to the load inductance, otherwise load current could not be continuous. But I am not sure whether the 2 diodes that form half of the bridge are usually referred to as freewheeling diodes, though. I'll have to look at this more closely.

It's the diode directly across the load that I think of as the classic freewheeling diode.
 
I tried with ur concept of FWD and got this answer as:
upload_2016-12-1_18-58-36.png

upload_2016-12-1_18-59-33.png


Thus i get the book answer.
Book has also given another explanation. They have taken fwd as diode D1 and D2.
They said that from
1) from alpha to pie, current flows through T1, load, D2
2) from pie to pie + alpha, current flows from T1 load, D1
So D1 is in freewheeling mode only in second case. Duration is pie + alpha - pie which is alpha. This happens in each half cycle of sine wave, so twice in full cycle.
Fraction is 2 * alpha divide by 2 * pie which gives alpha/pie.

But i like first explanation of mine better where ur concept of FWD is used. Thanks..
 

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